COMEDK 2025 Evening Shift | Circular Motion Question 1 | Physics

A point marked on a ring of radius 2 cm is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X - direction. Then the angle made by the displacement vector of the point with the X - axis is:

$\theta=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\mathbf{3} \boldsymbol{\pi}}\right)$

$\boldsymbol{\theta}=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\boldsymbol{\pi}}\right)$

$\theta=\tan ^{-1}\left(\frac{2 \pi}{3}\right)$

$\theta=\cot ^{-1}\left(\frac{2}{\pi}\right)$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

-0

A coin is placed on a disc rotating with an angular velocity $\omega$. The co-efficient of friction between the disc and the coin is $\mu$. The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is

$\sqrt{\frac{\mu}{\omega^2}}$

$\frac{\mu g}{\omega^2}$

$\sqrt{\frac{\mu g}{\omega^2}}$

$\frac{\mu g}{\omega}$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

A ball is moving in a circular path of radius $$5 \mathrm{~m}$$. If tangential acceleration at any instant is $$10 \mathrm{~ms}^{-2}$$ and the net acceleration makes an angle of $$30^{\circ}$$ with the centripetal acceleration, then, the instantaneous speed is

$$5.4 \mathrm{~ms}^{-1}$$

$$50 \sqrt{3} \mathrm{~ms}^{-1}$$

$$6.6 \mathrm{~ms}^{-1}$$

$$9.3 \mathrm{~ms}^{-1}$$

COMEDK 2024 Morning Shift

MCQ (Single Correct Answer)

-0

A metal ball of $$20 \mathrm{~g}$$ is projected at an angle $$30^{\circ}$$ with the horizontal with an initial velocity $$10 \mathrm{~ms}^{-1}$$. If the mass and angle of projection are doubled keeping the initial velocity the same, the ratio of the maximum height attained in the former to the latter case is :

1 : 2

2 : 1

1 : 3

3 : 1

COMEDK Subjects